Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $y = \dfrac{a^2 + 4a}{-a^2 - 5a - 4} \times \dfrac{2a^2 + 20a + 18}{-a^2 + 3a} $
Explanation: First factor out any common factors. $y = \dfrac{a(a + 4)}{-(a^2 + 5a + 4)} \times \dfrac{2(a^2 + 10a + 9)}{-a(a - 3)} $ Then factor the quadratic expressions. $y = \dfrac {a(a + 4)} {-(a + 1)(a + 4)} \times \dfrac {2(a + 1)(a + 9)} {-a(a - 3)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac {a(a + 4) \times 2(a + 1)(a + 9) } { -(a + 1)(a + 4) \times -a(a - 3)} $ $y = \dfrac {2a(a + 1)(a + 9)(a + 4)} {a(a + 1)(a + 4)(a - 3)} $ Notice that $(a + 1)$ and $(a + 4)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac {2a\cancel{(a + 1)}(a + 9)(a + 4)} {a\cancel{(a + 1)}(a + 4)(a - 3)} $ We are dividing by $a + 1$ , so $a + 1 \neq 0$ Therefore, $a \neq -1$ $y = \dfrac {2a\cancel{(a + 1)}(a + 9)\cancel{(a + 4)}} {a\cancel{(a + 1)}\cancel{(a + 4)}(a - 3)} $ We are dividing by $a + 4$ , so $a + 4 \neq 0$ Therefore, $a \neq -4$ $y = \dfrac {2a(a + 9)} {a(a - 3)} $ $ y = \dfrac{2(a + 9)}{a - 3}; a \neq -1; a \neq -4 $